Question

Let $\mathit{S}$be the subset of $\mathit{M}\mathbf{\left(}\mathbf{R}\mathbf{\right)}$consisting of all matrices of the form $\mathbf{\left(}\begin{array}{cc}\mathbf{a}& \mathbf{a}\\ \mathbf{b}& \mathbf{b}\end{array}\mathbf{\right)}$

1. Prove that $\mathit{S}$is a ring.
2. Show that $\mathit{J}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{1}& \mathbf{1}\\ \mathbf{0}& \mathbf{0}\end{array}\mathbf{\right)}$is a right identity in${\mathit{\mathbb{Z}}}_{\mathbf{n}}$ (meaning that $\mathit{A}\mathit{J}\mathbf{=}\mathit{A}$for every $\mathit{A}$in $\mathit{S}$).
3. Show that $\mathit{J}$ is not a left identity in $\mathit{S}$ by finding a matrix in role="math" localid="1659159033766" $\mathit{S}$ such that$\mathit{J}\mathit{B}\mathbf{\ne }\mathit{B}$ .

Short answer

(a) It is proved that S is a ring.

(b) It is proved that $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\in S$ is the right identity in S.

(c) It is proved that J is not the left identity in S.

Step-by-step solution

(a) Step 1: Prove that S is a ring

It is given that $S=\left\{\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)|a,b\in R\right\}\subseteq M\left(R\right)$. Assume that $\left(\begin{array}{cc}a& a\\ b& b\end{array}\right),\left(\begin{array}{cc}c& c\\ d& d\end{array}\right)\in S$.

Show the properties as:

(i) S is closed under addition.

$\begin{array}{c}\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)+\left(\begin{array}{cc}c& c\\ d& d\end{array}\right)=\left(\begin{array}{cc}a+c& a+c\\ b+d& b+d\end{array}\right)\\ \in S\end{array}$

(ii) S is closed under multiplication

$\begin{array}{c}\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)\left(\begin{array}{cc}c& c\\ d& d\end{array}\right)=\left(\begin{array}{cc}ac+ad& ac+ad\\ bc+bd& bc+bd\end{array}\right)\\ \in S\end{array}$

(iii) $0_R\in S$

$\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\in S$

(iv) If $a\in S$, then the solution of the equation$a+x=0_R$ lies in S.

$\begin{array}{c}\left(\begin{array}{cc}-a& -a\\ -b& -b\end{array}\right)\in S\\ \left(\begin{array}{cc}a& a\\ b& b\end{array}\right)+\left(\begin{array}{cc}-a& -a\\ -b& -b\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\\ =0_{M\left(R\right)}\end{array}$

Since all the properties are satisfied, this implies that S is a subring of $M\left(R\right)$, that is, S is a ring.

(b) Step 2: Show that J is right identity 

Assume that $\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)\in S$.

Find right identity as:

$\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)$

As , $\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}a& a\\ b& b\end{array}\right)$then by the definition of right identity $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\in S$ is the right identity in S.

Hence, it is proved.

(c) Step 3: Show that J is right identity 

Assume that $\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\in S$.

Find right identity as:

$\begin{array}{c}\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}2& 2\\ 0& 0\end{array}\right)\\ \ne \left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\end{array}$

As $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\ne \left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$, then by the definition of left right identity $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\in S$ is not the left identity in S.

Hence, it is proved.